2. 4 OTHER SOURCES OF POTENTIAL ENERGY
2.9 FREE-BODY DIAGRAM METHOD
the spring, as shown in Figure 2.31(b). Equal and opposite again, the spring force is acting againstthe body. Let xrepresent the displacement of the particle to which the spring is attached. If the spring force is drawn for a positive value of x, it is labeled kx and is drawn acting away from the body. Now if the spring is in compression, xtakes on a negative value.
If the spring force is drawn actingawayfrom the body and xis negative, it is actually acting against the body as shown in Figure 2.31(c). Thus, the spring force is always drawn in the direction opposite to the that of positive displacement of the point to which it is attached.
Then the direction of the spring force always takes care of itself.
The force from a viscous damper always opposes the direction of motion of the point to which it is attached on a FBD of a SDOF system. Ifxrepresents the displacement of the particle to which a viscous damper is attached, then its velocity is . The force from the vis- cous damper drawn on the FBD opposes the direction of positive . If the velocity of the particle is in the opposite direction and is negative, it is the same situation shown Figure 2.32(c) where a negative force on a FBD is actually in the opposite direction. Thus, the force from a viscous damper always opposes the direction of positive motion of the par- ticle to which it is attached. Like the spring force, the direction always takes care of itself.
When the effective force diagram is drawn, the effective forces are drawn to be consis- tent with the positive direction of the generalized coordinates.
x# x# x#
(a)
(b) cx·
cx·
(c)
= |cx·|
x.
x.
x
c m
FIGURE 2.32
The sign of the viscous-damping force takes care of itself if it is drawn to the opposite of the positive motion of the point to which the viscous damper is attached.
E X A M P L E 2 . 1 6
The block of Figure 2.33(a) slides on a frictionless surface. Derive the differential equation governing the motion of the system using xas the displacement of the system from its equi- librium position and as the generalized coordinate.
S O L U T I O N
The free-body diagram of Figure 2.33(b) shows the forces acting on the block at an arbi- trary instant. The spring force is kxand is drawn away from the block, indicating the spring is in tension for a positive x. The damping force is labeled and is drawn opposite the pos- itive direction of motion.
cx#
Applying Newton’s law to the free-body diagram in the xdirection leads to
(a) Rearranging the equation so that all terms involving the generalized coordinate are on one side yields
(b) Equation (b) is the governing differential equation. The values of x(0) and must be specified before solving.
x# (0) mx$
+cx#
+kx = F(t) -kx-c x#
+ F(t) = mx$
(a)
(b) F(t) kx
cx·
mg m c
F(t) k
FIGURE 2.33
(a) System of Example 2.16. Mass-spring and viscous-damper system sliding on a frictionless surface with an external force.
A thin disk of mass moment of inertia Iis attached to a fixed shaft of length L. The polar moment of inertia of the shaft is Jand it is made from a material of shear modulus G, as shown in Figure 2.34(a). A moment M(t) is applied to the disk. Derive the differential equation governing the clockwise angular displacement of the disk .
S O L U T I O N
The effect of the shaft is to produce a resisting moment
(a) on the disk. The disk undergoes pure rotational motion about the axis of the shaft. A FBD of the disk at an arbitrary time is shown in Figure 2.34(b). Applying to the disk and noting that leads to
(b)
Iu (c)
$ +
JG
L u = M(t) -
JG
L u + M(t) = Iu
$ a = u
$ πMG =Ia
M = JG
L u
u
E X A M P L E 2 . 1 7
θ M(t)
M(t) JG
L (a)
(b) FIGURE 2.34
(a) System of Example 2.17.
The angular displacement of the disk is the chosen gen- eralized coordinate. (b) FBD of the system at an arbitrary instant.
u
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E X A M P L E 2 . 1 8
The system of Figure 2.35 lies in a horizontal plane on a frictionless surface. Derive the dif- ferential equation governing the displacement of the mass.
S O L U T I O N
Let xrepresent the displacement of the mass. The disks move together. Assuming the cable connecting the block to the disk is inextensible, the change in length of the cable is x, which must be the amount of cable taken up or let out by the disk. If represents the clockwise angular rotation of the disk, the amount of cable let out is equal to the arc length subtended by as
(a) Equation (a) is valid for all time. It can be differentiated leading to and This is consistent with use of the relative velocity and relative acceleration equations applied between the center of the disk and the point instantaneously releasing the cable.
The acceleration of the point also has a component equal to directed toward the center of rotation. Using the same principle, the spring is stretched by 2x.
ru
#
2
x$
= ru
$ . x#
= ru
# x = ru
u
u
2r r
m k
I
(a)
= F(t)
R
2kx
External forces (b) F(t)
mx¨
m
I
Effective forces (c)
x¨r
FIGURE 2.35
(a) System of Example 2.18 lies in a horizontal plane. (b) FBDs of the system at an arbitrary instant. The system consists of the disk and the block.
FBD’s illustrating the external forces for the system and the effective forces are shown in Figure 2.35(b). Applying to these FBDs yields
(b) which is rearranged to
a1 (c)
r + mrbx$
+ 4krx = rF(t) -k(2x)(2r) + rF(t) = Iax$
rb + mx$ (r) (πMO)ext = (πMO )eff
E X A M P L E 2 . 1 9
A thin disk of mass mand radius r, has a spring of stiffness k, and has a viscous damper of damping coefficient cattached at its mass center, as shown in Figure 2.36(a). The disk rolls without slipping. Derive a differential equation governing the displacement of the mass center.
S O L U T I O N
Let xbe the displacement of the disk’s mass center. When the disk rolls without slipping the friction force is less than the maximum available friction force where Nis the normal force. The point of contact between the disk and the surface has a velocity of zero.
Use of the relative velocity equation between the point of contact and the center of mass yields
(a) The mass center only has a velocity and an acceleration in the horizontal direction; thus, Equation (a) can be differentiated to yield
(b) When the disk rolls without slipping, the kinematic condition of Equation (b) exists between the disk’s angular acceleration and the acceleration of the mass center. Noting that
a = ra
v = vC + vG>C = rvi
mN I = 1
2 mr2,
F kx + cx.
mx·. mr2
=
External forces Effective forces mg
No slip Thin disk of mass m and radius r c
µ k
N
1 2
x·.
r FIGURE 2.36
(a) System of Example 2.19. The disk rolls without slipping. (b) FBDs of the system at an arbitrary instant. The friction force is less than the maxi- mum available friction, and a kine- matic relationship exists between the angular acceleration and the acceler- ation of the mass center.
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FBDs of the disk at an arbitrary instant are shown in Figure 2.36(b). Summing moments on these FBDs according to leads to
(c) 3 (d)
2mx$ + cx#
+ kx = 0 -kx(r)-cx#
(r)= 1
2mr2ax$ rb+mx$
(r)
(πMC)ext =(πMC)eff a = x$
,
E X A M P L E 2 . 2 0
An accelerometer used in micro-electromechanical (MEMS) applications is shown in Figure 2.37(a). The accelerometer consists of a rigid bar between two massless fixed-fixed beams that are acting like springs. The bar is free to vibrate in the surrounding medium, which provides viscous damping. Derive a differential equation for the free vibrations of the accelerometer using a one degree-of-freedom model.
S O L U T I O N
The system is modeled, as in Figure 2.37(b), as a rigid bar attached to two identical springs.
The mass of the bar is
(a)
(b)
(c) beams
200 µm
200µm mass
Top view Cross-section of mass
Cross-section of beams
Side view
Direction of vibration
silicone
E = 1.9 × 1011 N/m2
20 mm
0.5 mm
1 mm 0.5 mm
c k
m
h1 = 15 mm
h2 = 10 mm
FIGURE 2.37
(a) MEMS accelerometer consists of a rigid bar between two fixed-fixed beams which vibrates in a viscous liquid. (b) SDOF model of system.
(c) Calculation of viscous damping coefficient.
(a) The moment of inertia of the cross section of one beam is
(b) The equivalent stiffness is twice the stiffness of a fixed-fixed beam at its midspan. From Appendix D, it is calculated as
(c)
An equivalent viscous-damping coefficient is calculated using an approximate linear veloc- ity profile in the surrounding fluid. The fluid on the top and bottom of the beam is in motion due to the vibrations of the beam as shown in Figure 2.37(c). The fluid above the beam has a velocity profile of
(d) where yis a coordinate into the fluid from the fixed surface. The shear stress acting on the beam is calculated using Newton’s viscosity law as
(e) and the resultant force on the surface of the beam is
(f) Using a similar analysis, the force on the lower surface of the beam is
(g) The total damping force is expressed as
(h) from which the equivalent viscous damping coefficient is calculated as
a 1
15 * 10-6 m + 1 10 * 10-6 mb
= (740 * 10-6N
#
s/m)(200 * 10-6 m)(20 * 10-6 m) ceq = mLda1h1 + 1 h2b F = mLda1
h1 + 1 h2bv F2 = mLd v
h2 F1 = tLd = mLd v
h1 t = mdu
dy = mv h1 u(y) =
v h1y
= 2192(1.9 * 1011N/m2)(4.17 * 10-26m4)
(200 * 10-6m)3 = 0.380N/m keq = 2a192EI
L3 b I =
1 12th3 =
1
12(0.5 * 10-6)(1.0 * 10-6m)3 = 4.17 * 10-26 m4
* (200 * 10-6m) = 4.6 * 10-12kg
= a2.3 g
cm3b a100cm
m b3 a 1kg
1000gb(20*10-6m)(0.5 * 10-6) meq = rdtL
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(i) The mathematical model for the free response of the system is
(j)